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Principles of Microeconomics 微观经济学代写|Statistics统计代写 - ECMT1010 Introduction to Economic Statistics

时间：2020-10-12

11. What is the sample correlation between the number of students accepted and enrolled?A)−0.83B)−0.24C) 0D) 0.24E) 0.83Use the following to answer questions 12–17.Your friend Donald is often late for meetings. To provea point, you want to studyμ, the mean time Donald is late for meetings. The following table contains theminutes by which Donald is late for 7 random meetings (a negative value means Donald was early).101370−58382712. What is ̄x, the mean time that Donald is late?A) 13B) 20C) 21D) 23E) 2713. To study how accurately ̄xapproximatesμ, you decide to use the bootstrap. Which of the followingis a valid bootstrap sample?A){38, 38, 38, 38, 38, 38, 38}B){10, 13, 70, 8, 8,−5, 70, 8}C){27, 27, 10, 10,−5,−6, 13}D){1, 2, 3, 4, 5, 6, 7}E){38, 70, 13, 10, 8,−5}14. Using StatKey, you obtain the following bootstrap distribution of sample means:Using the 95% rule, what is the 95% confidence interval forμ?A) (−0.3, 57.3)B) (4.1, 46.6)C) (5.5, 40.4)D) (9.6, 38.1)E) (14.3, 31.7)Page 4 of 13
15. Which of the following is most likely the 99% confidence interval forμ?A) (−0.3, 57.3)B) (4.1, 46.6)C) (5.5, 40.4)D) (9.6, 38.1)E) (14.3, 31.7)16. Which of the following is most likely the 90% confidence interval forμ?A) (−0.3, 57.3)B) (4.1, 46.6)C) (5.5, 40.4)D) (9.6, 38.1)E) (14.3, 31.7)17. To get a more accurate estimate ofμ, you want to have a much lower standard error for the bootstrapdistribution. Which of the following can help you achieve this?A) Increase the number of bootstrap samples to 10,000.B) Use a more powerful computer for bootstrapping.C) Measure and record the times Donald is late for work for a year and bootstrap on this largersample.D) Get Donald to randomly meet his wife 30 times, record the time he is late and bootstrap onthis larger sample.E) Randomly meet Donald 30 more times, record the times he is late and bootstrap on this largersample.Use the following to answer questions 18–21.LetAandBbe two events such thatP(A)=0.35,P(B)=0.45 andP(AandB)=0.1575.18. FindP(notA).A) 0.35B) 0.45C) 0.55D) 0.65E) 0.8519. FindP(AorB).A) 0.1000B) 0.1575C) 0.6425D) 0.8000E) 0.9575Page 5 of 13
20. FindP(B|A).A) 0.1000B) 0.1575C) 0.3663D) 0.4500E) 1.228621. Which of the following description about eventsAandBare correct?A)AandBare complementary.B)AandBare mutually exclusive and independent.C)AandBare NOT mutually exclusive and NOT independent.D)AandBare mutually exclusive but NOT independent.E)AandBare independent but NOT mutually exclusive.Use the following to answer questions 22–25.There are three roofing companies that service a smallcommunity. Al’s Roof Repair gets 45% of the roofing jobs in the community while Bob’s Better Building andCarl’s Roof Service get 25% and 30% of the business, respectively. Of Al’s customers, 70% are satisfied.Of Bob’s customers, 95% are satisfied. Among Carl’s customers, 90% are satisfied.22. What is the probability that a randomly selected customer used Al’s Roof Repair and is not satisfied?A) 0.135B) 0.165C) 0.300D) 0.315E) 0.66723. What is the probability that a randomly selected customer used Bob’s Better Building and is satisfied?A) 0.0125B) 0.2375C) 0.2661D) 0.7125E) 0.950024. What proportion of roofing customers are satisfied?A) 0.1775B) 0.5500C) 0.8225D) 1.0000E) 2.550025. If a randomly selected customer is satisfied, what is the probability that they used Al’s Roof Repair?A) 0.3150B) 0.3830C) 0.4500D) 0.5471E) 0.8511Page 6 of 13
Use the following to answer questions 26–30.Tony, the owner of a small auto repair store, wants toopen a second store in Penrith, but will do so only if more than half of Penrith’s households have cars(otherwise there won’t be enough business). He sampled 300 households in Penrith and found that 165have cars. Letpbe the population proportion of car-owning households in Penrith.26. What are the null and alternative hypotheses that summarize Tony’s decision?A)H0:p=0,Ha:p>0B)H0:p=0,Ha:p>0.5C)H0:ˆp=0,Ha:ˆp>0.5D)H0:ˆp=0.5,Ha:ˆp>0.5E)H0:p=0.5,Ha:p>0.527. What are the consequences of Type I and Type II errors in this case?Type I errorType II errorA)Don’t open a 2nd store; miss out on a goodbusiness opportunityOpen a 2nd store; get insufficient businessB)Don’t open a 2nd store; avoid a bad businessmistakeOpen a 2nd store; get insufficient businessC)Open a 2nd store; get insufficient businessOpen a 2nd store; reap the benefits of agood businessD)Open a 2nd store; get insufficient businessDon’t open a 2nd store; miss out on a goodbusiness opportunityE)Don’t open a 2nd store; avoid a bad businessmistakeDon’t open a 2nd store; miss out on a goodbusiness opportunity28. Tony obtained a bootstrap distribution of sample proportions of car-owning households in Penrithwith a standard error of 0.03. What is the 95% confidence interval forp?A) (0.42, 0.48)B) (0.48, 0.62)C) (0.49, 0.61)D) (0.52, 0.58)E) It cannot be determined with the given information.29. How can Tony generate a randomization distribution in this situation?A) Mark each of the 300 answers (with or without car) on individual cards. Shuffle the cardsand drawwith replacement300 times. Find the proportion of times getting a ‘with cars’ card.Repeat 10,000 times.B) Mark each of the 300 answers (with or without car) on individual cards. Shuffle the cards anddrawwithout replacement300 times. Find the proportion of times getting a ‘with cars’ card.Repeat 10,000 times.C) Ask the same 300 households the same question again on another day, and record the propor-tion of having cars. Repeat 10,000 times.D) Ask a different group of 300 households the same question, and record the proportion of havingcars. Repeat 10,000 times.E) Toss a fair coin 300 times independently and find the proportion of tosses with a head facingup. Repeat 10,000 times.Page 7 of 13
30. A randomization distribution under the null hypothesis is shown below:Tony solicited the help of some friends to form a conclusion. Their answers are as follows.Andrew:“Reject at the 5% significance level.”Eric:“Reject at the 2% significance level.”Joe:“Reject at the 1% significance level.”Based on the sample and the randomization distribution, who made the correct decision?A) They are all wrong.B) Only Andrew is correct.C) Only Eric is correct.D) Only Joe is correct.E) Only Eric and Joe are correct.Page 8 of 13
Long Question 1[20 marks total—suggested time approx. 44 minutes]Using data on 24 inkjet printers for sale on The Good Guys website in April 2017, a researcher obtainsthe following regression output for a model used to predict printer price (measured in dollars) based onprinting speed (measured in pages per minute, PPM).Predictor Coefficient SE coef. t statIntercept 28.5714 46.2082 0.6183PPM 6.0757 2.1028 2.8893Regression statisticsR square 0.2751 SD error 74.3863 Observations 24Analysis of varianceSource df SSRegression 1 46,192.22Residual 22 121,733.12Total 23 167,925.33a) Determine the correlation between pages per minute and price.[2 marks]b) Test whether the correlation between pages per minute and price is statistically significant at the1% level. Show all your steps.[3 marks]c) State in words the conclusion from the correlation test of significance.[2 marks]d) Give an interpretation of the slope coefficient.[2 marks]e) Test whether the slope coefficient is statistically significant at the 1% level. Show all your steps.[3marks]The researcher uses the bootstrap to investigate the regression slope estimate. The following shows theresults from 1,000 bootstrap samples.f) Briefly explain the purpose of the bootstrap distribution in this context.[2 marks]g) Use the bootstrap distribution to build a 99% confidence interval for the slope parameter.[2marks]h) Comment on the findings in b), e), and g).[2 marks]i) What do you think about the overall validity of this study? Explain.[2 marks]Page 9 of 13
Long Question 2[20 marks total—suggested time approx. 44 minutes]The unemployment rate is the proportion of adults in the population who are without work and are activelyseeking employment. According to the Australian Bureau of Statistics, the unemployment rate was 6% inOctober 2015. Suppose you have employment information on a randomly-selected sample of 50 adults inOctober 2015. Suppose the employment statuses of these 50 people are independent of each other. LetXbe the number of individuals in the sample who were without work and actively seeking employment.a) What is the distribution ofX? (You must specify all the parameters of the distribution.)[2 marks]b) What is the expected value ofX?[1 mark]c) What is the probability that the sample unemployment rate exactly equals the population unem-ployment rate? Round your answer to 3 decimal places.[2 marks]d) What is the probability that more than 3 adults in the sample are unemployed? Round your answerto 3 decimal places.[3 marks]In order to reduce the length of spells of unemployment, the government introduces a new program.To test the impact of the program, 7 unemployed individuals are randomly selected to undergo the newprogram and their employment status is tracked over time. A further 7 randomly-selected unemployedindividuals not participating in the program also have their employment status tracked. The table belowcontains the length of unemployment spells (measured in days) for the two groups.Program (treatment group)67818573707886No program (control group)75868777768186e) Suppose you want to test whether the program reduces the length of unemployment spells. Clearlydefine your notation and state the null and alternative hypotheses.[2 marks]f) Complete the test in (e) and clearly state the conclusion.[3 marks]g) Now suppose instead of being randomly selected, each of the 7 people in the control group waschosen so that they matched closely with someone in the treatment group. Thus, each person in thetreatment group now has a close match with a person in the control group (i.e., same age, gender,education, training, skill, location and so forth). Clearly define your notation and state the null andalternative hypotheses to test whether the program reduces the length of unemployment spells inthis case.[2 marks]h) Complete the test in (g) and clearly state the conclusion.[3 marks]i) Why are the hypothesis test results in (f) and (h) so different? Which is a better way to collect thedata to answer the question of whether unemployment spells are reduced by the program? Explain.[2 marks]END OF EXAMINATIONPage 10 of 13
Formulas, definitions, and distribution tablesPopulation and sample statistics.StatisticPopulationSamplesizeNnmeanμ=∑Ni=1xiN ̄x=∑ni=1xinstandard deviationσ=√√√∑Ni=1(xi−μ)2Ns=√√√∑ni=1(xi− ̄x)2n−1correlationρ=1NN∑i=1(xi−μx)σx(yi−μy)σyr=1n−1n∑i=1(xi− ̄x)sx(yi− ̄y)syDescriptive statistics.StatisticDefinitionz-scorezi=xi− ̄xsrangerange=max−mininter-quartile rangeIQR=Q3−Q1outliersxiQ3+1.5(IQR)95% rule ̄x±2sinterval estimatestatistic±margin of error95% confidence intervalstatistic±2×S EStandard deviations and standard errors for various statistics.StatisticStandard deviationStandard error ̄xσpnspnˆp√√p(1−p)n√√ˆp(1−ˆp)n ̄x1− ̄x2√√√σ21n1+σ22n2√√√s21n1+s22n2 ̄x1− ̄x2√√√p1(1−p1)n1+p2(1−p2)n2√√√ˆp1(1−ˆp1)n1+ˆp2(1−ˆp2)n2Page 11 of 13
Confidence intervals.100(1−α)% confidence interval:statistic±z∗α/2×S EforN(0, 1)distributionstatistic±t∗df,α/2×S Efortdistribution with df=n−1Test statistics for various hypotheses.Null hypothesisTest statisticH0:μ=μ0 ̄x−μ0s/pn∼tn−1H0:p=p0ˆp−p0pp0(1−p0)/n∼N(0, 1)H0:μ1−μ2=0 ̄x1− ̄x2»s21n1+s22n2∼tn−1wheren=min(n1,n2)H0:p1−p2=0ˆp1−ˆp2rÄ1n1+1n2äˆp(1−ˆp)∼N(0, 1)whereˆp=x1+x2n1+n2Selected percentiles from theN(0, 1)distribution.Right-tail probabilityConfidence levelz∗0.10080%1.2820.05090%1.6450.02595%1.9600.01098%2.3260.00599%2.575Selected percentiles fromtdistributions with various degrees of freedom (df).Right-tail probabilitydf0.050.0250.010.00561.9432.4473.1433.70871.8952.3642.9973.499221.7172.0742.5082.819231.7142.0692.5002.807241.7112.0642.4922.797Probability rules.Conditional probability:P(A|B)=P(AandB)p(B)Multiplicative rule:P(AandB)=P(A|B)P(B)Independence:P(A|B)=P(A)Mutual exclusion:P(AandB)=0Page 12 of 13
Law of total probability.P(A)=P(AandB)+P(Aand(notB))P(A)=P(AandB1)+P(AandB2)+···+P(AandBk)where(B1,B2, . . . ,Bk)are disjointBayes’ rule for two cases.P(A|B)=P(B|A)P(A)P(B|A)P(A)+P(B|notA)P(notA)Bayes’ rule forj=1, 2, . . . ,k.P(Aj|B)=P(B|Aj)P(Aj)P(B|A1)P(A1)+P(B|A2)P(A2)+···+P(B|Ak)P(Ak)where(A1,A2, . . . ,Ak)are disjoint.Population statistics for a discrete random variableXwith probability functionp(x).Mean:μ=n∑i=1xip(xi)Standard deviation:σ=√√√n∑i=1(xi−μ)2p(xi)SupposeXfollows a binomial distribution with parametersnandp.Binomial probability:P(X=k)=Ånkãpk(1−p)n−k=n!k!(n−k)!pk(1−p)n−kExpected value:n×pStandard deviation:pnp(1−p)Simple linear regression.Population regression model:y=β0+β1x+εSample regression model:ˆy=b0+b1x100(1−α)% confidence interval forβkbk±t∗df,α/2×S Ebkfork=0, 1 with df=n−2tstatistic forH0:βk=0t=bkS Ebkfork=0, 1 with df=n−2tstatistic forH0:ρ=0t=rpn−2p1−r2with df=n−2Goodness-of-fit:R2=r2=SSRSSTStandard deviation of the error:sε=√√SSEn−2Page 13 of 13